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$X$ and $Y$ denote Hilbert spaces. If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism? Homeomorphism means continuous map with continuous inverse.

I think the answer is yes, the only thing I am unable to show is what is the inverse of $T^*$ and if it is continuous?

Julien
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matt.w
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    This is true, more generally, for $X,Y$ any normed vector spaces. Essentially, this is because $|T^|=|T|$ and $(ST)^=T^S^$. – Julien May 21 '13 at 18:29
  • @Julien I think X and Y need to be banach. i.e to be able to talk about the adjoint of the inverse. https://math.stackexchange.com/questions/3127388/if-t-in-bx-y-and-t-is-bijective-is-t-also-bijective/3127391#3127391 – Jhon Doe Feb 26 '19 at 13:01

2 Answers2

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Let $S$ be inverse to $T$. Then $ST=1_X$ and $TS=1_Y$. Apply to this equalities $^*$ functor to get $T^*S^*=1_{X^*}$ and $S^*T^*=1_{Y^*}$. This means that $T^*$ is invertible and what is more $(T^*)^{-1}=S^*$

In this proof I implicitly assumed that $T^*\in \mathcal{B}(Y^*,X^*)$ whenever $T\in\mathcal{B}(X,Y)$. This is indeed true. For the begining recall one of the corollaries of Hahn-Banach theorem $$ \Vert g\Vert=\sup\{|g(y)|:y\in\operatorname{Ball}_{Y}\} $$ where $g\in Y^*$. Then $$ \begin{align} \Vert T^*\Vert &=\sup\{\Vert T^*(g)\Vert: g\in\operatorname{Ball}_{X^*}\}\\ &=\sup\{\Vert T^*(g)(x)\Vert: g\in\operatorname{Ball}_{Y^*}, x\in\operatorname{Ball}_{X}\}=\\ &=\sup\{| g(T(x))|: g\in\operatorname{Ball}_{Y^*}, x\in\operatorname{Ball}_{X}\}=\\ &=\sup\{\Vert T(x)\Vert: x\in\operatorname{Ball}_{X}\}=\\ &=\Vert T\Vert \end{align} $$

Norbert
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  • Hi this proof seems to hold for all normed spaces but in this link https://math.stackexchange.com/questions/403509/adjoint-operator-bijective?noredirect=1&lq=1 you mentioned that the spaces have to be banach. Im probably confused. Could you clarify? – Jhon Doe Feb 26 '19 at 09:35
  • Because that link speaks about equivalence, not a one sided implication – Norbert Feb 26 '19 at 10:05
  • Thanks. Secondly, Doesn't that link give a counter example for the above statement. In the link isn't A also a linear and bounded function. – Jhon Doe Feb 26 '19 at 10:14
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The proof of boundedness of $T^\ast$:

$$\|T^\ast(x)\|^2 = \langle T^\ast x, T^\ast x\rangle= \langle TT^\ast x, x\rangle \leq \|TT^\ast x\|\|x\| \leq \|T\|\|T^\ast x\|\|x\|,$$ where the second to last inequality is Cauchy-Schwarz. If $\|T^\ast x\| = 0$, then there's nothing much to say. Otherwise, we divide and get $$\|T^\ast(x)\| \leq \|T\|\|x\|.$$ So $\|T^\ast\|$ is bounded and $\|T^\ast \| \leq \|T\|.$

The reason that $T^\ast$ is a bijection is given in other answsers. The inverse of $T^\ast$ is bounded since $(T^\ast)^{-1}$ is the adjoint of a continuous function, namely $T^{-1}$. (Or if you like shiny tools, you can use the open mapping theorem!)

Edit: Once you know about reflexivity of Banach spaces, the above will give $\|T^\ast\| \leq \|T^{\ast\ast}\| = \|T\|$ and so $\|T^\ast\| = \|T\|$ follows.

Zach L.
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